Answer:
The current in the second wire is 4.4 A.
Explanation:
Given that,
Distance =7.0 cm
Current in first wire = 2.0 A
The magnetic field strength is zero at distance of 2.2 cm from the first wire.
We need to calculate the current in the second wire
Using formula of magnetic field
[tex]B=B_{1}-B_{2}[/tex]
[tex]0=\dfrac{\mu_{0}I_{1}}{2\pi r_{1}}-\dfrac{\mu_{0}I_{2}}{2\pi r_{2}}[/tex]
[tex]\dfrac{\mu_{0}I_{1}}{2\pi r_{1}}=\dfrac{\mu_{0}I_{2}}{2\pi r_{2}}[/tex]
[tex]\dfrac{I_{1}}{r_{1}}=\dfrac{I_{2}}{(d-r_{1})}[/tex]
Here, [tex]r_{2}=d-r_{1}[/tex]
[tex]I_{2}=\dfrac{I_{1}\times(d-r_{1})}{r_{1}}[/tex]
Put the value into the formula
[tex]I_{2}=\dfrac{2.0\times(7.0-2.2)}{2.2}[/tex]
[tex]I_{2}=4.4\ A[/tex]
Hence, The current in the second wire is 4.4 A.
